For what values of $k$ does this system of equations have a unique solution? Here's my system of equations:
$\begin{cases}y + 2kz = 0\\x + 2y + 6z = 2\\kx + 2z = 1\\
\end{cases}$
So I have $\left[\begin{array}{ccc|c}1&2&6&2\\0&1&2k&0\\k&0&2&1\end{array}\right]$
When I row reduce, I get:
$\left[\begin{array}{ccc|c}1&0&6-4k&2\\0&1&2k&0\\0&0&2-6k-4k^2&1-2k\end{array}\right]$
Not really sure where to go from here...
Any suggestions much appreciated!
Thanks,
Mariogs
 A: While DonAntonio's answer is certainly correct, it is likely that your question comes from a class where determinants have not yet been discussed, so you may need a different perspective.
In that case, recall that your system will be inconsistent if, after row reduction, you have a row of the form
$( 0 \  0 \ 0 \mid 1)$ since this row would correspond to the equation $0x+0y+0z=1$ which clearly has no solutions.
On the other hand, you also don't want a row of of the form $( 0 \ 0 \ 0 \mid 0)$, which would give you a free variable and hence infinitely many solutions.
Thus, you should find the values of $k$ for which $2-6k-4k^2 = 0$. By our discussion, we can see that as long as you avoid those values of $k$, your system will have a solution, and this solution will be unique.
You can find these "bad" values of $k$ by methods from high school algebra, e.g. the quadratic formula.
A: Now the last row could be rewritten as 
$$\pmatrix{1&0&6-4k&2\\0&1&2k&0\\ 0 & 0&2(2k-1)(k-1)&-(2k-1)}$$ 
Now we have the following cases


*

*$k=1$: the system has no solution.
$$\pmatrix{1&0&2&2\\0&1&2&0\\ 0 & 0&0&-1}$$ 

*$k=1/2$: the system has infinitely many solution.
$$\pmatrix{1&0&4&2\\0&1&1&0\\ 0 & 0&0&0}$$ 

*$k\in \mathbb{R}, k\ne 1,1/2$: the system has unique solution.

A: The coefficients' matrix of your non-homogeneous linear system is:
$$\begin{pmatrix}0&1&2k\\
1&2&6\\
k&0&2\end{pmatrix}$$
and the system has (a unique, by the way) solution iff the above matrix is regular, which happens iff its determinant is non-zero ...
